UNIT 6. Structured Query Language (SQL) Text: Chapter 5

Transcription

1 UNIT 6 Structured Query Language (SQL) Text: Chapter 5

2 Learning Goals Given a database (a set of tables ) you will be able to express a query in SQL, involving set operators, subqueries and aggregations rewrite SQL queries in one style (with one set of operators) with queries in a different style (using another set of operators) show that two SQL queries (with or without null values) are/aren t equivalent translate RA (or Datalog) queries to SQL queries and vice versa write SQL statements to insert, delete, update the database and define views write SQL statements to set certain constraints (in the project) use JDBC and Java to design DB transactions for the database users Unit 6 2

3 Outline Data Definition Language Basic Structure Set Operations Aggregate Functions Null Values Nested Subqueries Modification of the Database Views Integrity Constraints Embedded SQL, JDBC Unit 6 3

4 The SQL Query Language Developed by IBM (System R) in the 1970s Often pronounced as SEQUEL! Need for a standard since relational queries are used by many vendors Standards: SQL-86 SQL-89 (minor revision) SQL-92 (major revision, current standard) SQL-99 (major extensions) Consists of several parts: Data Definition Language (DDL) Data Manipulation Language (DML) o o Data Query Data Modification Unit 6 4

5 Creating Tables in SQL(DDL) A SQL relation schema is defined using the create table command: create table r (A 1 D 1, A 2 D 2,..., A n D n, (integrity-constraint 1 ),..., (integrity-constraint k )) Integrity constraints (ICs) can be: primary and candidate keys foreign keys general assertions o e.g., check (grade between 0 and 100) Example: CREATE TABLE Student CHAR(20) not null, (cid name CHAR(20), address CHAR(20), phone CHAR(8), major CHAR(4), primary key (cid)) Unit 6 5

6 Domain Types in SQL char(n). Fixed length character string with length n. varchar(n). Variable length character strings, with maximum length n. int. Integer (machine-dependent). smallint. Small integer (machine-dependent). numeric(p,d). Fixed point number, with user-specified precision of p digits, with d digits to the right of decimal point. real, double precision. Floating point and double-precision floating point numbers, with machine-dependent precision. float(n). Floating point number, with user-specified precision of at least n digits. Null values are allowed in all the domain types. To preclude null values declare attribute to be not null create domain in SQL-92 and 99 creates user-defined domain types e.g., create domain person-name char(20) not null Unit 6 6

7 Date/Time Types in SQL date. Dates, containing a (4 digit) year, month and date E.g. date time. Time of day, in hours, minutes and seconds. E.g. time 09:00:30 time 09:00:30.75 timestamp: date plus time of day E.g. timestamp :00:30.75 Interval: period of time E.g. Interval 1 day Subtracting a date/time/timestamp value from another gives an interval value Interval values can be added to date/time/timestamp values Relational DBMS offer a variety of functions to extract values of individual fields from date/time/timestamp convert strings to dates and vice versa For instance in Oracle (date is a timestamp): o TO_CHAR( date, format) o TO_DATE( string, format) o format looks like: DD-Mon-YY HH:MI.SS Unit 6 7

8 Running Examples Customer Database Customer(cid: integer, cname: string, rating: integer, salary: real) Item(iid: integer, iname: string, type: string) Order(cid: integer, iid: integer, day:date, qty:real) Student Database Student (sid, name, address, phone, major) Course (dept, cno, title, credits) Instructor( iname, degree) Section (dept, cno, secno, term, ins_name) Enrolled (sid, dept, cno, secno, term, mark) Prerequisite (dept, cno, pre_dept, pre_cno) Unit 6 8

9 Basic SQL Query SQL is based on set and relational operations with certain modifications and enhancements A typical SQL query has the form: select [distinct] A 1, A 2,..., A n from r 1, r 2,..., r m where P A i s represent attributes r i s represent relations P is a predicate. This query is nearly equivalent to the relational algebra expression. A1, A2,..., An ( P (r 1 x r 2 x... x r m )) The result of a SQL query is a table (relation). If distinct is used, duplicates are eliminated. By default duplicates are not eliminated! RA s projection RA s selection RA s join is done explicitly in P Dup-elim is expensive. How would you implement it? Unit 6 9

10 Basic SQL Query Called conjunctive query. Equivalent RA expression involves select, project, and join. So, also called SPJ query. SPJ/conjunctive queries correspond in Datalog to: p X r 1 Y,, r k Z, V i op U j,, V l op c, Union of SPJ queries (i.e., SPJU) queries => set of Datalog rules with the same head p X. SPJ with aggregation => SPJA queries; no counterpart in (pure datalog), although researchers have extended Datalog with aggregation (we won t cover this). Unit 6 10

11 Conceptual Evaluation Strategy Typical SQL query: SELECT FROM relation-list WHERE qualification [DISTINCT] attr-list Semantics of a SQL query defined in terms of the following conceptual evaluation strategy (in order): Compute the cross-product of relation-list. Discard any resulting tuples that fail qualifications. Drop attributes that are not in attr-list. If DISTINCT is specified, eliminate duplicate rows. This strategy is not a super efficient way to compute a query! An optimizer will find more efficient strategies to compute the same answers. Unit 6 11

12 Example Instances We will use these instances of the Customer, Item and Order relations in our examples. Customer cid cname rating salary 40 J. Justin G. Grumpy R. Rusty Order cid iid day qty /10/ /12/06 5 Item iid iname type 100 Inspiron6400 laptop 102 LatituteD520 laptop 105 DimensionE520 desktop 110 CanonMP830 printer Unit 6 12

13 Example of Conceptual Evaluation SELECT cname FROM Customer, Order WHERE Customer.cid=Order.cid AND iid=105 Customer X Order (cid) cname rating salary (cid) iid day qty 40 J. Justin /10/ J. Justin /12/ G. Grumpy /10/ G. Grumpy /12/ R. Rusty /10/ R. Rusty /12/06 5 Unit

14 Renaming Attributes in Result SQL allows renaming relations and attributes using the as clause: old-name as new-name Example: Find the name of customers who have ordered item 105 and the day they placed the order; rename cname to customer_name : SELECT cname AS customer_name, day FROM Customer, Order WHERE Customer.cid=Order.cid AND iid=105 Unit 6 14

15 Range Variables We can use variables to name relations in the FROM clause Usually used when same relation appears twice. The previous query can also be written as: OR SELECT cname, day FROM Customer C, Order R WHERE C.cid=R.cid AND iid=105 SELECT C.cname, R.day FROM Customer C, Order R WHERE C.cid=R.cid AND R.iid=105 Nothing but ans N, D customer I, N, R, S, order I, `105, D, Q. Unit 6 15

16 Using DISTINCT Find customers (id s) who ve ordered at least one item : SELECT C.cid FROM Customer C, Order R WHERE C.cid=R.cid Would adding DISTINCT to this query make a difference? Suppose we replace C.cid by C.cname in the SELECT clause. Would adding DISTINCT to this variant of the query make a difference? What if we use * in SELECT (* selects whole tuples)? SELECT * FROM Customer C, Order R WHERE C.cid=R.cid Unit 6 16

17 Expressions and Strings SELECT C.salary, tax=(c.salary-10)*0.3, C.salary*0.01 AS prof_ fees FROM Customer C WHERE C.cname LIKE B_%B Illustrates use of arithmetic expressions and string pattern matching: Returns triples of values, each consisting of the salary the income tax deducted (30% of salary minus 10K) and the professional fees (1% of the salary) for customers whose names begin and end with B and contain at least three characters. Unit 6 17 AS and = are two ways to name fields in result.

18 More on Strings LIKE is used for string matching: `_ stands for any one character and `% stands for 0 or more arbitrary characters. To match the name Strange%, need to use an escape character: like Strange\% escape \ SQL supports a variety of string operations such as concatenation (using ) converting from upper to lower case (and vice versa) finding string length, extracting substrings, etc. Unit

19 Ordering of Tuples List in alphabetic order the names of the customers who have ordered a laptop select cname from Customer, Item, Order where Customer.cid = Order.cid and Item.iid= Order.iid and type= laptop order by cname Order is specified by: Ordering goes beyond RA and Datalog! desc for descending order or asc for ascending order; ascending order is the default. E.g., order by cname desc Unit 6 19

20 Set Operations union, intersect, and except operate on tables (relations) and correspond to the RA operations Each of the above operations automatically eliminates duplicates; To retain all duplicates use the corresponding multiset versions: union all, intersect all and except all. Suppose a tuple occurs m times in r and n times in s, then, it occurs: m + n times in r union all s min(m,n) times in r intersect all s max(0, m n) times in r except all s Unit 6 20

21 Set Operations : UNION What do we get If we replace OR by AND here? Example: Find cid s of customers who ve ordered a laptop or a desktop SELECT C.cid FROM Customer C, Item I, Order R WHERE C.cid=R.cid AND R.iid=I.iid AND (I.type= laptop OR I.type= desktop ) UNION can be used to compute the union of any two compatible (corresponding attributes have same domains) sets of tuples (which are themselves the result of SQL queries): SELECT C.cid FROM Customer C, Item I, Order R WHERE C.cid=R.cid AND R.iid=I.iid AND I.type= laptop UNION SELECT C.cid FROM Customer C, Item I, Order R WHERE C.cid=R.cid AND R.iid=I.iid AND I.type= desktop Unit 6 21

22 Set Operations : EXCEPT EXCEPT can be used to compute the difference of two compatible sets of tuples Some systems use MINUS instead of EXCEPT What does the following query return? SELECT C.cid FROM Customer C, Item I, Order R WHERE C.cid=R.cid AND R.iid=I.iid AND I.type= laptop EXCEPT SELECT C.cid FROM Customer C, Item I, Order R WHERE C.cid=R.cid AND R.iid=I.iid AND I.type= desktop What does EXCEPT remind you of in Datalog? Unit 6 22

23 Set Operations: INTERSECT Example: Find cid s of customers who ve ordered a laptop and a desktop item : SELECT C.cid FROM Customer C, Item I1, Order R1, Item I2, Order R2 WHERE C.cid=R1.cid AND R1.iid=I1.iid AND C.cid=R2.cid AND R2.iid=I2.iid AND I1.type= laptop AND I2.type= desktop ) INTERSECT can be used to compute the intersection of two compatible sets of tuples (included in SQL/92, but some systems may not support it). SELECT C.cid FROM Customer C, Item I, Order R WHERE C.cid=R.cid AND R.iid=I.iid AND I.type= laptop INTERSECT SELECT S.cid FROM Customer S, Item I, Order R WHERE C.cid=R.cid AND R.iid=I.iid AND I.type= desktop Important to include the Key! Unit 6 23

24 Nested Queries Find names of customers who ve ordered item #105: SELECT C.cname FROM Customer C WHERE C.cid IN (SELECT R.cid FROM Order R WHERE iid=105) A very powerful feature of SQL: a WHERE clause can itself contain a SQL query! (Actually, so can FROM and HAVING clauses.) To find customers who ve not ordered item #105, use NOT IN. To understand semantics of nested queries, think of a nested loops evaluation: For each Customer tuple, check the qualification by computing the subquery. Unit

25 Nested Queries with Correlation Find names of customers who ve ordered item #105: SELECT C.cname FROM Customer C WHERE EXISTS (SELECT * FROM Order R WHERE iid=105 AND C.cid=R.cid) EXISTS is another set operator: returns true if the set is not empty. UNIQUE checks for duplicate tuples: returns true if there are no duplicates. If UNIQUE is used above, and * is replaced by iid, finds customers with at most one order for item #105. (* denotes all attributes. Why do we have to replace * by iid?) Ilustrates why, in general, subquery must be re-computed for each Customer tuple. Unit 6 25

26 More on Set-Comparison Operators We ve already seen IN, EXISTS and UNIQUE. Can also use NOT IN, NOT EXISTS and NOT UNIQUE. Also available: op ANY, op ALL, where op is one of: >, <, =, <=, >=, <> Find customers whose salary is greater than that of every customer with last name Rusty : SELECT * FROM Customer C How did we write such queries in RA? How about Datalog? WHERE C.salary > ALL (SELECT C2.salary FROM Customer C2 WHERE C2.cname LIKE % Rusty ) Unit a

27 Rewriting INTERSECT Queries Using IN Find cid s of customers who ve ordered both a laptop and a desktop: SELECT C.cid FROM Customer C, Item I, Order R WHERE C.cid=R.cid AND R.iid=I.iid AND I.type= laptop AND C.cid IN (SELECT C2.cid FROM Customer C2, Item I2, Order R2 WHERE C2.cid=R2.cid AND R2.iid=I2.iid AND I2.type= desktop ) Similarly, EXCEPT queries can be re-written using NOT IN. To find names (not cid s) of customers who ve ordered both laptops and desktops, just replace C.cid by C.cname in SELECT clause. Could we replace cid by cname throughout? Unit (What 6 about INTERSECT query?) 27

28 Division in SQL Find customers who ve ordered all items. Let s do it the hard way without EXCEPT: (1) select customer C such that... there is no item I (2) which is not ordered by C SELECT cname FROM Customer C WHERE NOT EXISTS ((SELECT I.iid FROM Item I) EXCEPT (SELECT R.iid FROM Order R WHERE R.cid=C.cid)) SELECT cname FROM Customer C WHERE NOT EXISTS (SELECT * FROM Item I WHERE NOT EXISTS (SELECT * FROM Order R WHERE R.iid=I.iid AND R.cid=C.cid)) How does it compare with RA/Datalog? Unit

29 Aggregate Operators These functions operate on the multiset of values of a column of a relation, and return a value AVG: average value MIN: minimum value MAX: maximum value SUM: sum of values COUNT: number of values The following versions eliminate duplicates before apply the operation to attribute A: COUNT ( DISTINCT A) SUM ( DISTINCT A) AVG ( DISTINCT A) Unit 6 29

30 Aggregate Operators: Examples SELECT COUNT (*) FROM Customer SELECT AVG (salary) FROM Customer WHERE rating=10 SELECT cname FROM Customer C WHERE C.rating= (SELECT MAX(C2.rating) FROM Customer C2) SELECT AVG ( DISTINCT salary) FROM Customer WHERE rating=10 SELECT COUNT (DISTINCT rating) FROM Customer WHERE salary BETWEEN 50 AND 100 Unit 6 30

31 Aggregate Operators: Examples(cont) Find name and salary of the richest customer(s) The first query is wrong! WHY? Second query is fine: can use value = subquery only if subquery returns single value. The third query is equivalent to the second query, and is allowed in the SQL/92 standard, but is not supported in some systems. SELECT cname, MAX (salary) FROM Customer SELECT cname, salary FROM Customer WHERE salary = (SELECT MAX (salary) FROM Customer) SELECT cname, salary FROM Customer WHERE (SELECT MAX (salary) FROM Customer) = salary Unit 6 31

32 GROUP BY and HAVING Often, we want to divide tuples into groups and apply aggregate operations to each group. Example: Find the average salary of the customers in each rating level. Suppose we know that rating values go from 1 to 10; we can write 10 queries that look like this (!): Problem: For i = 1, 2,..., 10: SELECT AVG (salary) FROM Customer WHERE rating = i We don t know how many rating levels exist, and what the rating values for these levels are! Solution: Use GROUP BY and/or HAVING clauses Unit 6 32

33 GROUP BY and HAVING (cont) SELECT [DISTINCT] target-list FROM relation-list WHERE qualification GROUP BY grouping-list HAVING group-qualification ORDER BY target-list The target-list contains (i) attribute names (ii) terms with aggregate operations (e.g., AVG (salary)). Attributes in (i) must also be in grouping-list. each answer tuple corresponds to a group, a group is a set of tuples that have the same value for all attributes in grouping-list selected attributes in (i) must have a single value per group. Attributes in group-qualification are either in grouping-list or are arguments to an aggregate operator. Unit 6 33

34 Conceptual Evaluation of a Query 1. compute the cross-product of relation-list 2. keep only tuples that satisfy qualification 3. partition selected tuples into groups by the value of attributes in grouping-list 4. keep only the groups that satisfy group-qualification ( expressions in group-qualification must have a single value per group!) 5. keep only the fields that are in target-list 6. generate one answer tuple per qualifying group. Unit 6 34

35 GROUP BY Example Example1: For each item, find the salary of the poorest customer who has ordered this item: SELECT FROM WHERE GROUP BY iid iid, MIN (salary) Customer C, Order R C.cid= R.cid Unit 6 35

36 GROUP BY Example: Default Evaluation Customer cid cname rating salary 40 J. Justin G. Grumpy R. Rusty Order cid iid day qty /10/ /12/ /06/ /06/07 5 C.cid=R.cid (Customer X Order) (cid) cname rating salary (cid) iid day qty 40 J. Justin /10/ J. Justin /06/ G. Grumpy /06/ R. Rusty /12/06 5 Unit 6 36

37 GROUP BY Example (cont ) C.cid=R.cid (Customer X Order) and grouped by iid (cid) cname rating salary (cid) iid day qty 40 J. Justin /10/ J. Justin /06/ G. Grumpy /06/ R. Rusty /12/06 5 Result iid Min(salary) Unit , 5a

38 GROUP BY and HAVING Example Example2: For each item that has more than 2 orders, find the salary of the poorest customer who has ordered this item: SELECT iid, MIN (salary) FROM Customer C, Order R WHERE C.cid= R.cid GROUP BY iid HAVING COUNT (*) > 2 Unit

39 Grouping Examples (contd.) For each laptop item, find the number of (distinct) customers who ordered this item SELECT I.iid, COUNT (DISTINCT R.cid) AS scount FROM Item I, Order R WHERE R.iid=I.iid AND I.type= laptop GROUP BY I.iid Grouping over a join of two relations. What do we get if we: (a) remove I.type= laptop from the WHERE clause, and then (b) add a HAVING clause with this dropped condition? What if we replace COUNT (DISTINCT R.cid) with COUNT (*)? Unit 6 39

40 More Grouping Examples For each rating that has at least 2 customers whose salary is at least 50K, find the average salary of these customers for that rating SELECT rating, AVG (salary) FROM Customer WHERE salary >= 50 GROUP BY rating HAVING COUNT (*) > 1 Only rating can appear alone in the SELECT and/or HAVING clauses. 2nd column of result is unnamed. (Use AS to name it.) Customer cid cname rating salary 22 J. Justin R. Rubber Z. Zorba H. Hasty B. Brutus R. Rusty rating salary rating Answer relation Unit

41 Grouping Examples (cont ) For each rating that has at least 2 customers (of any salary), find the average salary among the customers with that rating whose salary is at least 50K. SELECT C.rating, AVG (C.salary) FROM Customer C WHERE C.salary >= 50 GROUP BY C.rating HAVING 1 < (SELECT COUNT (*) FROM Customer C2 WHERE C2.rating=C.rating) Shows HAVING clause can also contain a subquery. Compare this with the query where we concidered only ratings with at least 2 customers with salary at least 50K! What if HAVING clause is replaced by: HAVING COUNT(*) >1 Unit

42 Grouping Examples (contd.) Find those ratings for which their average salary is the minimum over all ratings SELECT C.rating FROM Customer C WHERE C.salary = (SELECT MIN (AVG (C2.salary)) FROM Customer C2) WRONG! Aggregate operations cannot be nested! Correct solution (in SQL/92 and SQL/99 ) SELECT Temp.rating, Temp.avgsalary FROM (SELECT C.rating, AVG (C.salary) AS avgsalary FROM Customer C GROUP BY C.rating) AS Temp WHERE Temp.avgsalary = (SELECT MIN (Temp.avgsalary) FROM Temp) Unit 6 42

43 Null Values Tuples may have a null value, denoted by null, for some of their attributes Value null signifies an unknown value or that a value does not exist. The predicate IS NULL ( IS NOT NULL ) can be used to check for null values. E.g., Find the names of the customers whose salary is not known. SELECT cname FROM Customer WHERE salary IS NULL The result of any arithmetic expression involving null is null E.g., 5 + null returns null. Unit 6 43

44 Null Values and Three-Valued Logic To deal with null values we need a three-valued logic using the truth value unknown: OR False Unknown True False F U T Unknown U U T T True T T T U Classical Logic is 2-valued. F Lattice of truth values in Kleene s 3-valued logic. Unit 6 44

45 Null Values and Three-Valued Logic To deal with null values we need a three-valued logic using the truth value unknown: AND False Unknown True False F F F Unknown F U U T True F U T U Classical Logic is 2-valued. F Lattice of truth values in Kleene s 3-valued logic. Unit 6 45

46 Null Values and Three-Valued Logic To deal with null values we need a three-valued logic using the truth value unknown: False Unknown True NOT T U F T U Classical Logic is 2-valued. F Lattice of truth values in Kleene s 3-valued logic. Unit 6 46

47 Null Values and Three-Valued Logic In particular, note: OR: (unknown or true) = true, (unknown or false) = unknown (unknown or unknown) = unknown AND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown NOT: (not unknown) = unknown P is unknown evaluates to true if predicate P evaluates to unknown Any comparison with null returns unknown E.g. 5 < null or null <> null or null = null Result of where clause predicate is treated as false if predicate evaluates to unknown All aggregate operations except count(*) ignore tuples with null values on the aggregated attributes. Unit 6 47

48 Summary of impact of null values Represent unknown or inapplicable. Make arithmetic expressions evaluate to unknown (U). Make logical expressions evaluate to T, U, of F. SQL treats truth value U in where clause as F. SQL ignores tuples with nulls on aggregated attr when aggregating using any function save Count. SQL counts all tuples regardless of presence of nulls. SQL lets you test if a value is null via IS NULL and IS NOT NULL. Unit 6 48

49 Database Modification Insertion Can insert a single tuple using: INSERT INTO Student VALUES (53688, Smith, 222 W.15 th ave, , MATH) or INSERT INTO Student (sid, name, address, phone, major) VALUES (53688, Smith, 222 W.15 th ave, , MATH) Add a tuple to student with null address and phone: INSERT INTO Student (sid, name, address, phone, major) VALUES (33388, Chan, null, null, CPSC) What if there was a not null constraint on address or phone? Unit 6 49

50 Database Modification Insertion (contd.) Can add values selected from another table? Add an order for customer 222 for every laptop item with date 1/1/07 and quantity 5 INSERT INTO Order SELECT 222, iid, 1/1/07, 5 FROM Item WHERE type = laptop Query-driven insert. The select-from-where statement is fully evaluated before any of its results are inserted So, statements like INSERT INTO table1 SELECT... FROM table1 are ok. How would you say add orders for a specific item (iid) from every customer rated at 6 or above? Unit 6 50

51 Database Modification Deletion Note that only whole tuples are deleted. Can delete all tuples satisfying some condition (e.g., name = Smith): DELETE FROM Student WHERE name = Smith Delete all Customers whose salary is above the average sailor salary: DELETE FROM Customer WHERE salary > (SELECT avg(salary) FROM Customer) Do you see any problem with this? Unit 6 51

52 Database Modification Updates Increase the rating of all customers by 2 (should not be more than 10) Need to write two updates: UPDATE Customer SET rating = 10 WHERE rating >= 8 UPDATE Customer SET rating = rating + 2 WHERE rating < 8 Is the order important? How would you raise by 10% the salary of every customer rated at 8 or above? Alternatively, use case statement: UPDATE Customer SET rating = CASE WHEN rating< 8 THEN rating+2 WHEN rating >= 8 THEN 10 END ELSE rating Unit 6 52

53 Views Provide a mechanism to hide certain data from certain users. To create a view we use the command: where: CREATE VIEW vname AS <query expression> <query expression> is any legal SQL expression vname is the view name Example: CREATE VIEW LDOrder AS SELECT cid, iid, type, date FROM Item I, Order R WHERE I.iid = R.iid AND (type = laptop OR type = desktop ) Unit

54 With Clause Allows views to be defined locally to a query, rather than globally. Example: WITH LDOrder(cid, iid, type, date) AS SELECT cid, iid, type, date FROM Item I, Order R WHERE I.iid = R.iid AND (type = laptop OR type = desktop ) SELECT cid, cname FROM Customer C, LDOrder R WHERE C.cid = R.cid AND date > 1/1/07 Unit 6 54

55 SQL s Join Operators Join operations take two relations and return as a result another relation. These additional operations are typically used as subquery expressions in the from clause Join condition defines which tuples in the two relations match, and what attributes are present in the result of the join. Join type defines how tuples in each relation that do not match any tuple in the other relation (based on the join condition) are treated. Example: Join Types inner join left outer join right outer join full outer join Join Conditions natural on <predicate> SELECT S.name, E.dept, E.cno FROM Student S NATURAL LEFT OUTER JOIN Enrolled E Unit

56 Integrity Constraints (Review) An IC describes conditions that every legal instance of a relation must satisfy. Inserts/deletes/updates that violate IC s are disallowed. Can be used to ensure application semantics (e.g., cid is a key), or prevent inconsistencies (e.g., cname has to be a string, salary must be < 200) Types of IC s: domain constraints, primary key constraints, foreign key constraints, general constraints Unit 6 56

57 General Constraints CREATE TABLE Customer Create with a CHECK clause. cid INTEGER, Constraints can be named cname CHAR(10), Can use subqueries to express constraint rating INTEGER, salary REAL, CREATE TABLE Order1 PRIMARY KEY (cid), ( cid INTEGER, CHECK ( rating >= 1 iid INTEGER, AND rating <= 10 ); day DATE, qty REAL, PRIMARY KEY (cid, iid, day), CHECK (`Printer <> ( SELECT I.type FROM Item I WHERE I.iid=iid))); Check constraints are checked when tuples are inserted or modified Unit 6 57

58 Domain Constraints User can create a new domain and set constrains for it. Example: CREATE DOMAIN agedomain INTEGER DEFAULT 21 CHECK ( VALUE >= 1 AND VALUE <= 110 ) In current systems distinct domains are not distinct types. New systems support distinct types Unit 6 58

59 Constraints Over Multiple Relations Cannot be defined in one table. Are defined as ASSERTIONs which are not associated with any one table. Example: Every student has taken at least one course. CREATE ASSERTION totalenrolment CHECK ( NOT EXISTS ((SELECT sid FROM student) EXCEPT (SELECT sid FROM Enrolled))); What would this IC look like in Datalog-like syntax? Unit 6 59

60 Transactions A transaction is a sequence of queries and update statements executed as a single unit Transactions are started implicitly and terminated by one of o commit work: makes all updates of the transaction permanent in the database o rollback work: undoes all updates performed by the transaction. Example Transfer of money from account A to account B involves two steps: o deduct from A and add to B If one step succeeds and the other fails, database is in an inconsistent state Therefore, either both steps should succeed or neither should If any step of a transaction fails, all work done by the transaction can be undone by rollback work. Rollback of incomplete transactions is done automatically, in case of system failures Unit 6 71

61 Transactions (Contd.) In most database systems, each SQL statement that executes successfully is automatically committed. Each transaction consists of only a single statement Automatic commit can usually be turned off, but how to do so depends on the database system Another option in SQL:1999: enclose statements within begin atomic end Unit 6 72

62 Summary SQL was an important factor in the early acceptance of the relational model; more natural than earlier, procedural query languages. Relationally complete; in fact, significantly more expressive power than relational algebra. Consists of a data definition, data manipulation and query language. Many alternative ways to write a query; optimizer looks for most efficient evaluation plan. Holy Grail: users don t have to care about efficiency, and relegate finding an efficient plan to QOzer. In practice, users need to be aware of how queries are optimized and evaluated for best results. Unit 6 73

63 Summary (Contd.) NULL for unknown field values brings many complications SQL allows specification of rich integrity constraints (and triggers) Embedded SQL allows execution within a host language; cursor mechanism allows retrieval of one record at a time APIs such as ODBC and JDBC introduce a layer of abstraction between application and DBMS Unit 6 74